All categories

4 years ago

Find the slope of the tangent line to f(x)=e4x2+5x at the point where x=0

1 answer:

3 0

We find the first derivative of the curve, and evaluate it at x=1 (which we will use to obtain the slope of the line tangent to the curve at (1,−1)). So we have a point on the tangent line: (x0,y0)=(1,−1) and the slope of the line tangent at that point: m=−13.

You might be interested in

Please solve #13-14

Serga [27]

Answer:

XZ = 44

KQ = 24

Step-by-step explanation:

XH is half of XZ, and XH is 22x so two times 22x will give you XZ.

XH times two equation - 2(22x)

XZ equation - 43x + 1

Set them equal

2(22x) = 43x + 1

44x = 43x + 1

x = 1

Put x in equation XZ

43(1) + 1 = 44

5 0

3 years ago

If x=7 and y=−5, evaluate the following expression:3x−y/2

CaHeK987 [17]

Answer:

( 3 ) ( 7 ) − ( $\frac{-5}{2}$ )

= 21 − ( $\frac{-5}{2}$ )

= $\frac{47}{2}$

Hope it helps

Please mark me as the brainliest

Thank you

4 0

3 years ago

What is the following product? 35 2

umka21 [38]

the answer is the first one

7 0

3 years ago

Math help Please!!!!

GalinKa [24]

Part A. What is the slope of a line that is perpendicular to a line whose equation is −2y=3x+7?

Rewrite the equation −2y=3x+7 in the form $y=-\dfrac{3}{2}x-\dfrac{7}{2}.$ Here the slope of the given line is $m_1=-\dfrac{3}{2}.$ If $m_2$ is the slope of perpendicular line, then

$m_1\cdot m_2=-1,\\ \\m_2=-\dfrac{1}{m_1}=\dfrac{2}{3}.$

Answer 1: $\dfrac{2}{3}$

Part B. The slope of the line y=−2x+3 is -2. Since $-\dfrac{3}{2}\neq -2\quad \text{and}\quad \dfrac{2}{3}\neq -2,$ then lines from part A are not parallel to line a.

Since $-2\cdot \left(-\dfrac{3}{2}\right)=3\neq -1\quad \text{and}\quad -2\cdot \dfrac{2}{3}=-\dfrac{4}{3}\neq -1,$ both lines are not perpendicular to line a.

Answer 2: Neither parallel nor perpendicular to line a

Part C. The line parallel to the line 2x+5y=10 has the equation 2x+5y=b. This line passes through the point (5,-4), then

2·5+5·(-4)=b,

10-20=b,

b=-10.

Answer 3: 2x+5y=-10.

Part D. The slope of the line $y=\dfrac{x}{4}+5$ is $\dfrac{1}{4}.$ Then the slope of perpendicular line is -4 and the equation of the perpendicular line is y=-4x+b. This line passes through the point (2,7), then